When 10 m L of an aqueous solution of F e2+ ions was titrated in the presence of dil H 2 SO 4 using diphenylamine indicator, 15 mL of 0.02 M solution of K 2 Cr 2 O 7 was required to get the end point. The molarity of the solution containing F e2+ ions is x × 10-2 M. The value of x is .(Nearest integer)

Question

When 10 m L of an aqueous solution of F e2+ ions was titrated in the presence of dil H 2 SO 4 using diphenylamine indicator, 15 mL of 0.02 M solution of K 2 Cr 2 O 7 was required to get the end point. The molarity of the solution containing F e2+ ions is x × 10-2 M. The value of x is .(Nearest integer) (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

milli-equivalents of F e2+= milli-equivalents of K 2 Cr 2 O 7 M × 10 × 1=0.02 × 15 × 6 M=0.18=18 × 10-2 M

milli-equivalents of Fe2+= milli-equivalents of K2​Cr2​O7​ M×10×1=0.02×15×6 M=0.18=18×10−2M

milli-equivalents of $F e^{2 +} =$ milli-equivalents of $K_{2} C r_{2} O_{7}$
$M \times 10 \times 1 = 0.02 \times 15 \times 6$
$M = 0.18 = 18 \times 1 0^{- 2} M$